Let $A$ be a unital Banach algebra and $f: Inv(A) \to A$ be the map $a \mapsto a^{-1}$. I'm trying to show that $f$ is differentiable. My idea is to show that the limit of $\delta \to 0$ of $$ {\|(a + \delta a)^{-1} - a^{-1}\| \over |\delta| \|a\|}$$ exists that is, is finite. I have now run out of ideas. Because I got stuck I showed instead that $f$ is continuous which was easy enough. I hoped to use it somehow to show that $f$ is also differentiable but no luck.
Please can someone explain to me how to show that $$ \lim_{\delta \to 0}{\|(a + \delta a)^{-1} - a^{-1}\| \over |\delta| \|a\|}$$ exists?
For $a \in \text{Inv}(A)$ let $$ L_a:A\to A,\quad L_a(h)=-a^{-1}ha^{-1}. $$ $L_a$ is obviously a linear continuous map, and for every $h \in A$ with $\|a^{-1}h\|<1$ we have $$ (a+h)^{-1}=(I+a^{-1}h)^{-1}a^{-1}=[I+\sum_{n=1}^\infty(-1)^n(a^{-1}h)^n]a^{-1}=a^{-1}+\sum_{n=1}^\infty(-1)^n(a^{-1}h)^na^{-1}. $$ It follows that \begin{eqnarray} \|f(a+h)-f(a)+a^{-1}ha^{-1}\|&\le&\sum_{n=2}^\infty\|(a^{-1}h)^na^{-1}\|\\ &\le&\|a^{-1}\|\sum_{n=2}^\infty\|a^{-1}h\|^n\\ &=&\|a^{-1}\|\frac{\|a^{-1}h\|^2}{1-\|a^{-1}h\|}. \end{eqnarray} Hence $$ \frac{\|f(a+h)-f(a)-L_a(h)\|}{\|h\|}\le\frac{\|a^{-1}\|^3\|h\|}{1-\|a^{-1}h\|}, $$ i.e. $f$ is differentiable and $$ df_a(h)=-a^{-1}ha^{-1} \quad \forall a \in \text{Inv}(A),\ h \in A. $$